Asked by Chasm
how many milliliters of a 0.46 M HCl solution are needed to react completely with a 6.4 g of zinc to form zinc (ii) chloride answer in mL
Help please :( I was absent and don't know where to start
Help please :( I was absent and don't know where to start
Answers
Answered by
Steve
start by reading your text book. I'm sure there's a section regarding moles.
Zn has a mol wt of 65.39, so 6.4g = 0.1 moles of Zn
Since the reaction is
Zn + 2HCl = ZnCl2 + H2
each mole of Zn reacts with 2 moles HCl
So, you need .2 moles HCl
.46M HCl has .46 moles/liter
So, you need .2/.46 = .434 liters, or 434ml
Zn has a mol wt of 65.39, so 6.4g = 0.1 moles of Zn
Since the reaction is
Zn + 2HCl = ZnCl2 + H2
each mole of Zn reacts with 2 moles HCl
So, you need .2 moles HCl
.46M HCl has .46 moles/liter
So, you need .2/.46 = .434 liters, or 434ml
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